... and beyond

How do I find a logistic function from its graph?

1 Answer

Hi there. A logistic graph is like an exponential with an upper limit, so it has two horizontal asymptotes, usually y=0 and y=B, as in the "spread of infection" graph here:

The curve is the solution to the diff eqn #dy/dt=ry(1-y/B)# with initial point #(t,y)=(0,y_0),#which can be solved by separation of variables and partial fractions! (Think of the starting point at the lower left.) The solution curve is given by

#y=(By_0)/(y_0 + (B-y_0)e^(-rt))#

If you have the graph, you can read off the #(0,y_0)#, the upper limit #B#, and the inflection point #(t_(".inflect"), B/2)#. (Infection point?)

The next part is to solve for #r# using the inflection point: that's where the second derivative is 0, so take the derivative of #dy/dt#
or the second derivative of the equation for y, and solve!

That part I'll leave for you. You're welcome, from \dansmath/ ;-}

[[Added comment from dansmath: I think the prevailing notation is #y=B/(1 + (B/y_0 - 1)e^(-rt)# which is the same equation, just divide top and bottom by #y_0#.]]